3. Calculate the theoretical weight force of the water (W) displaced by the metal cylinder at 1, 2, 3, 4, and 5cm. Wa = RavALag) where gy is the density of water (= 1.0 x10³ kg/m³) L, is the length of the cylinder submerged, and g is gravitational acceleration. Wa1 (L, 1cm) = _ W:3 (L. 3cm) = , Ws5 (L, 5cm) = W2 (L. 2cm) = . Ws4 (L. 4cm) = . Wubmerzed (Show one weight force calculation below. Be careful to convert to SI units.)

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1-please help with the highlighted question 3 (which depends on attached answers from part 2) for this buoyancy problem. Thank you!

2. Calculate the circular cross-sectional area (A) of your aluminum cylinder. (A = zR?)
_4.15*10^-4_m?
Diameter=2.3cm
R=2.3cm/2=1.15cm
R=1.15cm/100=0.0115m
A=n*R^2
A= n*(0.0115m)^2
A=0.000415|
3. Calculate the theoretical weight force of the water (W) displaced by the metal cylinder at
1, 2, 3, 4, and 5cm.
(Wa = RALag) where gy is the density of water (= 1.0 x103 kg/m³) L. is the length of the
cylinder submerged, and g is gravitational acceleration.
Ws1 (L. 1cm) =
Ws3 (L. 3cm) =
Ws5 (L. 5cm) =
Ws2 (Ls 2cm) =
W:4 (L. 4cm) =
Waubmerzed
%3D
(Show one weight force calculation below. Be careful to convert to SI units.)
Transcribed Image Text:2. Calculate the circular cross-sectional area (A) of your aluminum cylinder. (A = zR?) _4.15*10^-4_m? Diameter=2.3cm R=2.3cm/2=1.15cm R=1.15cm/100=0.0115m A=n*R^2 A= n*(0.0115m)^2 A=0.000415| 3. Calculate the theoretical weight force of the water (W) displaced by the metal cylinder at 1, 2, 3, 4, and 5cm. (Wa = RALag) where gy is the density of water (= 1.0 x103 kg/m³) L. is the length of the cylinder submerged, and g is gravitational acceleration. Ws1 (L. 1cm) = Ws3 (L. 3cm) = Ws5 (L. 5cm) = Ws2 (Ls 2cm) = W:4 (L. 4cm) = Waubmerzed %3D (Show one weight force calculation below. Be careful to convert to SI units.)
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